Talk:Kilogram

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This article is written in British English, which has its own spelling conventions (colour, travelled, centre, defence, artefact, analyse) and some terms may be different or absent from other varieties of English. According to the relevant style guide, this should not be changed without broad consensus.

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I was looking at reference 3 and it appears that this link no longer works. Can this be changed to https://link.springer.com/article/10.1007/s11018-019-01648-4 or does it need to be a PDF since springer is pay-walled for some.

I reverted this edit which changed the date format to mdy contrary to the style guide at MOS:DATERET. My revert was reverted with the edit summary The data format has been set since at least 2014. Please discuss in the Talk:Kilogram if you feel strongly this should change. I would prefer if we used the dmy format since it is a more international format used by most of the world. And I feel strongly about this case since this was the original format used in the article and was changed without a consensus. Thank you. Pynappel (talk) 18:52, 26 December 2023 (UTC)Reply

The mixture of UK spelling with US dates is weird. I would support a change to either UK spelling + UK date or US spelling + US date. I don't mind which. Dondervogel 2 (talk) 19:09, 26 December 2023 (UTC)Reply

My point is that per MOS:DATERET we should use the original date format, which is dmy, despite it having been changed in 2014 to mdy (without any discussion or consensus). Pynappel (talk) 10:59, 21 July 2024 (UTC)Reply

I have no objection to the change you propose. Perhaps we can then add a "this article uses dmy format" banner on the talk page? Dondervogel 2 (talk) 11:44, 21 July 2024 (UTC)Reply

I support the change to DMY format: this is an article with strong international ties and should use non-regional conventions, as well as consistency with international language variant in use. I see the purpose of MOS:DATERET as being only to avoid edit wars. However, even MOS:DATERET permits changing the style through consensus on the talk page. So far, it looks like we can reach consensus for a change to DMY format. —Quondum 19:30, 14 October 2024 (UTC)Reply

Though not a very active discussion, I believe that this qualifies as talk-page consensus for a change to DMY format (motivation for the change with support and without dissent, other than the revert that wanted the discussion here, with plenty of opportunity for people to chime in). Accordingly, I'm making the change. —Quondum 22:36, 19 October 2024 (UTC)Reply

Does anyone know how to document DMY formally in the Talk Page banner? Dondervogel 2 (talk) 10:27, 20 October 2024 (UTC)Reply

The closest thing that I've found is to use {{consensus|This article uses dmy date format}} (wording could be changed). —Quondum 13:34, 20 October 2024 (UTC)Reply

I found (and added) a banner for British English. That's a start. A separate one for DMY would be helpful. Dondervogel 2 (talk) 14:33, 20 October 2024 (UTC)Reply

The numeric date format should be YYYY-MM-DD, in accordance with ISO 8601. Formally, ISO 8601 does not necessarily apply to Wikipedia articles; but YYYY-MM-DD is the best format. Eschew surplusage. 94.30.84.71 (talk) 10:31, 27 July 2025 (UTC)Reply

I removed this definition of the kilogram: = 917 097 121 160 018 · 62 154 105 072 590 475^-1 · 10⁴² ⁠hΔν_Cs/c²⁠. The definition had been removed by @Remsense (semi-auto) after it was added by @Kamil Kielczewski then added back.

This form does not appear in the SI brochure so it should not be grouped with the other two. I started to move it to a separate paragraph where it could be marked unsourced, but then I realized I could not verify the content. As noted by Kamil Kielczewski in an edit summary the number is " is non-trivial to compute, " so editors will not verify it. Providing numerical values is not not a goal for Wikipedia, but verifable content is.

If you don't agree we can discuss this, but I assume Remsense will agree with my removal. Johnjbarton (talk) 17:02, 21 January 2025 (UTC)Reply

Perhaps I'm wrong here, but I also disagree with the reasons given for its reinstatement: we're an encyclopedia, not an algebra solver – we're here to give a well-rounded account of what sources have to say, not help folks do their calculations as such. Remsense ‥ 论 17:33, 21 January 2025 (UTC)Reply

Obtaining it is non-trivial, but verification is relatively simple, it only requires multiplication (if you have following factorization which is also easy to verify):

- numerator: (299792458)^2 = (2 × 7 × 73 × 293339)^2,

- denominator: 662607015 × 10^42 = 3 × 5 × 7 × 6310543 × 10^42 and 9192631770 = 2 × 3^2 × 5 × 7^2 × 47 × 44351

Thus, after canceling the common divisors, we have:

- numerator 917097121160018 = 2 × 73^2 × 293339^2 × 10^42

- denominator: 62154105072590475 = 3^3 × 5^2 × 7 × 47 × 44351 × 6310543

Alright, but let’s be consistent: the current approximate value ≈ 1.475521399735270 × 10^40 used in atricle (which can be easily obtained from the (removed) reduced form of the fraction, for example, here https://www.wolframalpha.com/input?i=100*917097121160018+%2F+62154105072590475 ) also must be removed and replaced with 1.47552134, as this is the value provided in the SI brochure you referenced. Kamil Kielczewski (talk) 17:50, 21 January 2025 (UTC)Reply

Thanks, I edited to match the source.

A good source on the other form and the reasons behind it would make all the difference for me. Verification is not proving the values are equal, but it is rather that "content is determined by previously published information rather than editors' beliefs, experiences". You think the other form is notable, but that is just an opinion. Johnjbarton (talk) 18:56, 21 January 2025 (UTC)Reply

I think you're touching on philosophy – whether a particular formula in a source article is important or true is also just the personal (or group) opinion of the author of that article... So, either way, we rely on someone's opinions... Kamil Kielczewski (talk) 00:06, 22 January 2025 (UTC)Reply

Weighing the importance and relevance of statements to the reader is what editing is. No need to mystify or problematize it, 's not like you can avoid doing so one way or the other. Remsense ‥ 论 00:10, 22 January 2025 (UTC)Reply

No, not philosophy, but a proven practice for encyclopedic knowledge. A published reliable source has been reviewed by multiple people and thus it is no longer a personal opinion. Reliable sources have authors, reviewers, and editors with track records. A group opinion of this kind is not similar to a personal opinion of an anonymous contributor. Johnjbarton (talk) 02:34, 22 January 2025 (UTC)Reply

Once, many people, even intelligent ones, believed that the Earth was the center of the universe, and Galileo was almost killed for his views... Just a thought on the subjectivity of group opinions. This applies to all modern theories and concepts, which are always subjective to individuals or groups. That's all from me on the philosophical topic of subjective opinions that you started – best regards! Kamil Kielczewski (talk) 08:04, 22 January 2025 (UTC)Reply

Yes, and if encyclopedias existed before Galileo, they would claim the Earth was the center of the universe as that was the best summary of the then current knowledge. An encyclopedia with a thousand people's opinions is worthless, even if Galileo's opinion is among them. There was no way to know his view will prevail. Johnjbarton (talk) 16:54, 22 January 2025 (UTC)Reply

By the way, when "canceling the common divisors", you missed two fives. The unique irreducible fraction is:

- numer.: 3 668 388 484 640 072 × 10^40 = 2^3 × 73^2 × 293339^2 × 10^40

- denom.: 2 486 164 202 903 619 = 3^3 × 7 × 47 × 44351 × 6310543

Birdfern (talk) Birdfern (talk) 20:51, 22 January 2025 (UTC)Reply

you can't "cancel" the factor 5^2 (from denominator) because the numerator (299792458^2) is not divisible by 5

${\displaystyle {\frac {299792458^{2}}{6.62607015\times 10^{-34}\times 9192631770}}={\frac {299792458^{2}}{662607015\times 9192631770}}\times 10^{42}=}$

${\displaystyle ={\frac {2^{2}\times 7^{2}\times 73^{2}\times 293339^{2}}{(3\times 5\times 7\times 6310543)\times (2\times 3^{2}\times 5\times 7^{2}\times 47\times 44351)}}\times 10^{42}={\frac {2^{2}\times 7^{2}\times 73^{2}\times 293339^{2}}{2\times 3^{3}\times 5^{2}\times 7^{3}\times 47\times 44351\times 6310543}}\times 10^{42}}$

${\displaystyle ={\frac {2\times 73^{2}\times 293339^{2}}{3^{3}\times 5^{2}\times 7\times 47\times 44351\times 6310543}}\times 10^{42}={\frac {917097121160018}{62154105072590475}}\times 10^{42}}$

This is elementary (although a bit time-consuming) mathematics, easy to verify - no sources needed.

Kamil Kielczewski (talk) 12:14, 23 January 2025 (UTC)Reply

10⁴² is divisible by 5. The fraction reduces to 3668388484640072×10⁴⁰ / 2486164202903619. Indefatigable (talk) 15:37, 23 January 2025 (UTC)Reply

And if one is not looking for the simplest fraction, but instead the shortest string that evaluates to this value in some computerized notation, then the answer is probably something like "85654988e20^2/4972328405807238" (30 characters). Birdfern (talk) 19:32, 23 January 2025 (UTC)Reply

Ok now I see, your formula is even better because it is compatible with the approximate decimal expansion of this fraction.

${\displaystyle ...={\frac {2\times 73^{2}\times 293339^{2}}{3^{3}\times 5^{2}\times 7\times 47\times 44351\times 6310543}}\times 2^{2}\times 5^{2}\times 10^{40}}$

${\displaystyle ={\frac {2^{3}\times 73^{2}\times 293339^{2}}{3^{3}\times 7\times 47\times 44351\times 6310543}}\times 10^{40}={\frac {3668388484640072}{2486164202903619}}\times 10^{40}}$

Thank you. Kamil Kielczewski (talk) 21:21, 23 January 2025 (UTC)Reply

Short definition of kilogram 2409:4089:A280:6FF7:10EE:6FA8:F759:1A83 (talk) 13:58, 5 March 2025 (UTC)Reply

It's in the lede and repeated later in the article: A kilogram is the mass of a liter of water. But this is not a formal definition. There is exactly one formal definition. It cannot be shortened and still be a definition. -Arch dude (talk) 14:48, 5 March 2025 (UTC)Reply